NP-hard problems of feasibility

Question

What does "deciding if a system X of equations has a solution is NP-hard" really mean?

Is it equivalent to say that finding a solution to this system is NP-hard?

If we can solve, we can decide. But what about the converse?

Thank you

Answer 1

No, it is not equivalent. For example, consider the system of equations $$ x_1 = 2, x_{2} = x_1^2, \ldots, x_n = x_{n-1}^2 $$ Deciding whether a solution exists is easy: the answer is yes. But that solution is $x_k = 2^{2^k}$, which has exponentially many bits, so you can't even write down the solution in polynomial time.